cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A040996 Maximum number of distinct functions at the bottom of a Boolean (or Binary) Decision Diagram (or BDD) with negation by pointer complementation.

Original entry on oeis.org

1, 6, 120, 32640, 2147450880, 9223372034707292160, 170141183460469231722463931679029329920, 57896044618658097711785492504343953926464851149359812787997104700240680714240
Offset: 0

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Comments

At 0, the last variable, the only choice is (t, f) because the first entry is always uncomplemented and the 2nd must be different.
At level 1, the 2nd-to-last variable, the first entry is either t or a pointer to a following level (0) and the 2nd entry is either of these or its negation, except it may not equal the first entry.
At level n, the n-th-to-last variable, the first entry is either t or a pointer to one of the following levels' functions and the second entry is any of these or its negation, but not equal to the first entry.
From Luis H. Gallardo, Nov 18 2021: (Start)
Another description of a(n) follows: let TP(n) = t^(2^n-1)*(t+1)^(2^n-1) in the ring F_2[t]. Expand TP(n) as a sum of monomials c*t^k in F_2[t], with c equal 0 or 1. Lift TP(n) to LTP(n) in the ring Z[t], i.e., consider the coefficients c of TP(n) to be integers in LTP(n), instead of elements of F_2. Finally, substitute t by 2 in LTP(n). We get: a(n) = LTP(n).
Example: a(3) = subs(t=2, TP(3)) = 32640, where TP(3) = t^14 + t^13 + t^12 + t^10 + t^9 + t^8 + t^7 = t^7*(t+1)^7 in F_2[t]. (End)

Crossrefs

Subsequence of A000217. Half of A111403.

Programs

  • Magma
    [2^(2^n)*(2^(2^n)-1)/2: n in [0..10]]; // Vincenzo Librandi, Sep 30 2011
  • Maple
    a(n) = subs(t=2,modp(expand(t^(2^n-1)*(t+1)^(2^n-1)),2)); # Luis H. Gallardo, Nov 18 2021
  • Mathematica
    f[x_]:=Module[{c=2^(2^x)},(c(c-1))/2]; Array[f,10,0] (* Harvey P. Dale, Sep 29 2011 *)
  • PARI
    a(n)=if(n<=0,n==0,2^(2^n)*(2^(2^n)-1)/2)
    

Formula

a(n) = (S(n-1) + 1) * (2*S(n-1) + 1) where S(n-1) = Sum_{k
a(n) is the (2^(2^n)-1)-th triangular number; i.e., a(n) = 2^(2^n)*(2^(2^n)-1)/2.
a(n) = A111403(n) / 2. - Tilman Piesk, Oct 04 2024